Question

In Fig, $△ODC\sim △OBA,\angle BOC={125}^o$ and $\angle CDO={70}^o$. Find $\angle DOC,\angle DCO$ and $\angle OAB$.

Answer 1

Since, $\angle COD+\angle COB={180}^o$
$\therefore$ $\angle COD={180}^o-{125}^o={55}^o$
Since, $\angle COD+\angle ODC+\angle DCO={180}^o$
$\therefore$ $\angle DCO={180}^o-{70}^o-{55}^o={55}^o$
Since, $\angle DCO=\angle OAB=$ Alternate angles
$\therefore$ $\angle OAB={55}^o$